F(x) = ∫[1,x] f(t) dt
= ∫[1,x] 2t+2 dt
= t^2+2t [1,x]
= (x^2+2x)-(1^2+2*1)
= x^2+2x-3
Use the graph of f(t) = 2t + 2 on the interval [–1, 4] to write the function F(x), where f of x equals the integral from 1 to x of f of t dt.
F(x) = x2 + 3x
F(x) = x2 + 2x – 12
F(x) = x2 + 2x – 3
F(x) = x2 + 4x – 8
2 answers
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